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Proof step 40 from proof datasize-further (previous step)

n + 1 = datasize(buf, n2), datasize(buf, n) = datasize(buf, n2) ↔ (∀ n1. n ≤ n1 ∧ ¬ n2 < n1 → buf[n1] = empty), buf[n2] = empty, n < n2, n < n0, n < # buf, datasize(buf, n2) ≤ n2, datasize(buf, n2) ≤ n0, datasize(buf, n2) ≤ # buf, n0 ≤ n2, n + 1 = datasize(buf, n2) ∧ (datasize(buf, n) = datasize(buf, n2) ↔ (∀ n3. n ≤ n3 ∧ n3 < n2 → buf[n3] = empty)) ∧ n < n0 ∧ n0 < n2 ∧ n < # buf ∧ n < n2 ∧ datasize(buf, n2) < n2 ∧ datasize(buf, n2) ≤ n0 ∧ datasize(buf, n2) ≤ # buf → buf[n0] = empty ∨ buf[n] = empty ⊦ buf[n0] = empty, buf[n] = empty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 41:
n + 1 = datasize(buf, n2), datasize(buf, n) = datasize(buf, n2) ↔ (∀ n1. n ≤ n1 ∧ ¬ n2 < n1 → buf[n1] = empty), buf[n2] = empty, n < # buf, n < n0, n < n2, n0 ≤ n2, datasize(buf, n2) ≤ # buf, datasize(buf, n2) ≤ n0, datasize(buf, n2) ≤ n2 ⊦ buf[n] = empty, buf[n0] = empty, (datasize(buf, n) = datasize(buf, n2) ↔ (∀ n3. n ≤ n3 ∧ n3 < n2 → buf[n3] = empty)) ∧ n0 < n2 ∧ datasize(buf, n2) < n2

Proof step 40 from proof datasize-further (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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